Decibel homework help

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  • #1
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They say that the unit bel was originally defined as:

[tex]B = \log\Big(\frac{P_1}{P_2}\Big) [/tex]

But then it turned out to be a large unit and people stated using decibels:

[tex]dB = 10\log\Big(\frac{P_1}{P_2}\Big) [/tex]

I don't get this. One decibel is one-tenth of a bel, then shouldn't dB be actually:

[tex]dB = \frac{1}{10}\log\Big(\frac{P_1}{P_2}\Big) [/tex]

?
 

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  • #2
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haha, good question. and somewhere along the lines, a dB is also 20log(something)
 
  • #3
berkeman
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I dunno about the "bel" thing, but dB are defined for power ratios and for voltage or current ratios.

Power in dBm is 10 log( P / 1mW )

Voltage in dBV is 20 log ( V / 1V )

Voltage in dBuV is 20 log ( V / 1uV )
 
  • #4
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there's nothing wrong with it. If you have 5 meters, then you have 10*5=50 decimeters, since there are 10 decimeters in each meter. :)

If you have 5 bels then you have 10*5=50 decibels, since there are 10 decibels in each bel. :)

The logs of the power ratio (Bels) just turned out to be too small and inconvenient. So instead they just decided to use tenths of a Bel, or decibels.

It's like measuring sewing yarn in kilometers. Not very convenient. You get very very small values as a result of the measurement using such a unit. Instead we multiply the result in kilometers by say 100,000, to get centimeters, which is a much easier number to deal with.
 
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  • #5
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haha, good question. and somewhere along the lines, a dB is also 20log(something)
No, a decibel is a measure of POWER RATIOS. It is RIGIDLY DEFINED as 1 dB = 10log (Po/Pi).

When you want to find the dB value of voltage or current ratios you plug P = V^2/R or P = I^2R into the above definition of the decibel. The Rs in the ratio cancel (if the voltages you are taking the ratio of are both across the same resistance) and the square dependency can be brought down in front of the log function yielding dB = 20log(Vo/Vi) or dB = 20log(Io/Ii).

In terms of powers it is 10log(Po/Pi) but in terms of voltage and current ratios it is 20log(Vo/Vi) or 20log(Io/Ii)
 
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  • #6
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there's nothing wrong with it. If you have 5 meters, then you have 10*5=50 decimeters, since there are 10 decimeters in each meter. :)

If you have 5 bels then you have 10*5=50 decibels, since there are 10 decibels in each bel. :)
What you have said makes a whole lot of sense. But what am I doing wrong?

Isn't 1bel = 10decibel then 1decibel = 1/10 bel and since 1 bel = log(P1/P2), then would wouldn't that make 1 decibel = 1/10*bel = 1/10*log(P1/P2).
 
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  • #7
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What you have said makes a whole lot of sense. But what am I doing wrong?

Isn't 1bel = 10decibel then 1decibel = 1/10 bel and since 1 bel = log(P1/P2), then would wouldn't that make 1 decibel = 1/10*bel = 1/10*log(P1/P2).
You did nothing wrong. 1 decibel = (1/10)*Bel, or 10 dBel = 1Bel. One decibel is one-tenth of a bel. However, to find the number of decibels in a bel you don't simply multiply the 1 bel by 1/10....you'd multiply by 10.

For instance, 1 meter is 1/1000 of a kilometer, but you don't multiply the number of kilometers by 1/1000 to get the number of meters! You multiply by 1000.
 
  • #8
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ok.. so 10 deciBel = 1Bel (ie, straightway we know the value in deciBel > value in Bel) so the dB must be dB = 10log(P1/P2), because deciBel value is 10 times bigger than the Bel value in this case.
 
  • #9
rbj
2,226
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They say that the unit bel was originally defined as:

[tex]B = \log\Big(\frac{P_1}{P_2}\Big) [/tex]

But then it turned out to be a large unit and people stated using decibels:

[tex]dB = 10\log\Big(\frac{P_1}{P_2}\Big) [/tex]

I don't get this. One decibel is one-tenth of a bel, then shouldn't dB be actually:

[tex]dB = \frac{1}{10}\log\Big(\frac{P_1}{P_2}\Big) [/tex]

?
when your unit is smaller, you need more of them to represent the same quantity. so scaling the number of dB up by a factor of 10 makes perfect sense if a dB is one tenth as large as a bel.

the root definition of the dB has to do with a "Just Noticable Difference" in loudness of a sound around 1 kHz for a human with typically good hearing (check out Fletcher-Munson curve). they could have defined it as

[tex]G_{\mathrm{dB}} = 3 \log_2 \left( \frac{P_1}{P_2} \right) [/tex]

instead of the base 10 log and, as someone who works with computers, audio, and music, i wish they did. it would make my numerical life a little easier.
 
  • #10
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DeciBel help

Here is the help you have been waiting for.

First lets look at a simple conversion:

Lets say that you want to convert 3 meters to decimeters. There are 2 basic ways that you should already know:

1. Moving the decimal point: 3.0m = 30.0dm = 300.0cm = 3000.0 mm , etc.
2. Use the factor label method common in Chemistry:

3m * (1dm/10^-1m) or 3m * (10dm/1m)

As you can see by the second example, to convert from meters to decimeters you multiply by 10 using the conversion factor of 10dm = 1m.

The same thing applies to the deciBel or dB:
The conversion factor is 1dB = 10^-1 B or equally 10dB = 1B

This is why you multiply by 10:

dB = 10 * log ( Power out/Power in)

MajorMath
 
  • #11
Averagesupernova
Science Advisor
Gold Member
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No, a decibel is a measure of POWER RATIOS. It is RIGIDLY DEFINED as 1 dB = 10log (Po/Pi).

When you want to find the dB value of voltage or current ratios you plug P = V^2/R or P = I^2R into the above definition of the decibel. The Rs in the ratio cancel (if the voltages you are taking the ratio of are both across the same resistance) and the square dependency can be brought down in front of the log function yielding dB = 20log(Vo/Vi) or dB = 20log(Io/Ii).

In terms of powers it is 10log(Po/Pi) but in terms of voltage and current ratios it is 20log(Vo/Vi) or 20log(Io/Ii)
How do you figure that 1 dB is 10log (Po/Pi)? If I have 10 watts in and 10 watts out that intuitively (as well as mathematically) is a zero dB change in power.
 
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  • #12
stewartcs
Science Advisor
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No, a decibel is a measure of POWER RATIOS. It is RIGIDLY DEFINED as 1 dB = 10log (Po/Pi).
No, the Gain in dB is equal to that (in an electronic sense).

CS
 
  • #13
1,306
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How do you figure that 1 dB is 10log (Po/Pi)? If I have 10 watts in and 10 watts out that intuitively (as well as mathematically) is a zero dB change in power.
I didn't mean to put the 1 in there. :) I meant dB = 10log(Po/Pi)

my bad. Hopefully my point is still clear to the OP despite that typo.....
 

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